/*
Search in a Binary Search Tree
==============================

You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:
Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:
Input: root = [4,2,7,1,3], val = 5
Output: []

Constraints:
The number of nodes in the tree is in the range [1, 5000].
1 <= Node.val <= 107
root is a binary search tree.
1 <= val <= 107
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution
{
public:
  TreeNode *searchBST(TreeNode *root, int val)
  {
    if (!root)
      return NULL;
    if (root->val == val)
      return root;
    else if (root->val > val)
      return searchBST(root->left, val);
    return searchBST(root->right, val);
  }
};
